The solution of the differential equation $\frac{dy}{dx}+\frac{2xy}{1+x^2}=\frac{1}{(1+x^2)^2}$ is: (A) $y(1+x^2)=c+\tan^{-1}x$ (B) $\frac{y}{1+x^2}=c+\tan^{-1}x$ (C) $y\log(1+x^2)=c+\tan^{-1}x$ (D) $y(1+x^2)=c+\sin^{-1}x$

> [!TLDR] > This is a first-order linear differential equation solved using the integrating factor method. The solution is $y(1+x^2) = c + \tan^{-1}x$, which corresponds to option **(A)**. The key to solving any first-order linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$ is to multiply both sides by an **integrating factor** — a function that turns the left-hand side into a perfect derivative. Once that happens, you can integrate directly. Here, the equation is: $$\frac{dy}{dx} + \frac{2xy}{1+x^2} = \frac{1}{(1+x^2)^2}$$ Let’s identify $P(x) = \frac{2x}{1+x^2}$ and $Q(x) = \frac{1}{(1+x^2)^2}$. 1. **Find the integrating factor (I.F.)** The formula is $\text{I.F.} = e^{\int P(x)\,dx}$. Compute $\int \frac{2x}{1+x^2}\,dx$. Notice that the numerator $2x$ is exactly the derivative of $1+x^2$. So: $$\int \frac{2x}{1+x^2}\,dx = \log|1+x^2|$$ Since $1+x^2 > 0$ for all real $x$, we drop the absolute value. Hence: $$\text{I.F.} = e^{\log(1+x^2)} = 1+x^2$$ 2. **Multiply the entire differential equation by the I.F.** $$(1+x^2)\frac{dy}{dx} + 2xy = \frac{1}{1+x^2}$$ The left-hand side is now exactly $\frac{d}{dx}\big[y(1+x^2)\big]$. Why? Because: $$\frac{d}{dx}\big[y(1+x^2)\big] = (1+x^2)\frac{dy}{dx} + y \cdot 2x$$ which matches perfectly. 3. **Rewrite and integrate** The equation becomes: $$\frac{d}{dx}\big[y(1+x^2)\big] = \frac{1}{1+x^2}$$ Integrate both sides with respect to $x$: $$y(1+x^2) = \int \frac{1}{1+x^2}\,dx = \tan^{-1}x + c$$ where $c$ is the constant of integration. …

The solution of the differential equation (dy)/(dx)+(2xy)/(1+x²)=(1)/((1+x²)²) is: (A) y(1+x²)=c+tan⁻¹x (B) (y

NCERT Exemplar · Q97

\frac{dy}{dx}+\frac{2xy}{1+x^2}=\frac{1}{(1+x^2)^2}

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Concept understanding — Integrating Factor Method

Integrating Factor Method

Some first-order differential equations refuse to separate — you cannot get all the $y$ 's on one side and all the $x$ 's on the other. The integrating factor method is the standard trick for a special (and very common) family of these: the linear first-order equation. The idea is beautifully simple: multiply the whole equation by one cleverly chosen function, and the messy left-hand side collapses into a single derivative that we can integrate directly.

The Standard Form

An equation is linear of first order if it can be written as

\frac{dy}{dx} + P\,y = Q

where $P$ and $Q$ are functions of $x$ alone (or constants). Notice $y$ and $\frac{dy}{dx}$ appear only to the first power, and never multiplied together — that is what "linear" means here.

Tip

Always rearrange into this exact shape first. The coefficient of $\frac{dy}{dx}$ must be $1$ before you read off $P$ and $Q$ .

The Integrating Factor

The magic multiplier is

\text{I.F.} = e^{\int P\,dx}.

Why this one? Multiply the equation by $e^{\int P\,dx}$ :

e^{\int P\,dx}\frac{dy}{dx} + P\,e^{\int P\,dx}\,y = Q\,e^{\int P\,dx}.

By the product rule, the left-hand side is exactly $\dfrac{d}{dx}\!\left(y\cdot e^{\int P\,dx}\right)$ , because the derivative of $e^{\int P\,dx}$ is $P\,e^{\int P\,dx}$ . So the equation becomes

\frac{d}{dx}\big(y\cdot \text{I.F.}\big) = Q\cdot \text{I.F.}

The left side is now a single derivative — that is the whole point of choosing this factor.

The Solution

Integrate both sides with respect to $x$ :

y\cdot \text{I.F.} = \int \big(Q\cdot \text{I.F.}\big)\,dx + C.

This is the general solution. In words: (solution) × (integrating factor) = integral of (Q × integrating factor), plus a constant.

A Quick Illustration

For $\dfrac{dy}{dx} + \dfrac{1}{x}\,y = x$ , we read $P = \tfrac1x$ , $Q = x$ . Then $\int P\,dx = \log x$ , so $\text{I.F.} = e^{\log x} = x$ . The solution is

y\cdot x = \int x\cdot x\,dx + C = \frac{x^3}{3} + C.

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This is a first-order linear differential equation solved using the integrating factor method. The solution is $y(1+x^2) = c + \tan^{-1}x$ , which corresponds to option (A).

The key to solving any first-order linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$ is to multiply both sides by an integrating factor — a function that turns the left-hand side into a perfect derivative. Once that happens, you can integrate directly.

Here, the equation is:

\frac{dy}{dx} + \frac{2xy}{1+x^2} = \frac{1}{(1+x^2)^2}

Let’s identify $P(x) = \frac{2x}{1+x^2}$ and $Q(x) = \frac{1}{(1+x^2)^2}$ .

Find the integrating factor (I.F.) The formula is $\text{I.F.} = e^{\int P(x)\,dx}$ . Compute $\int \frac{2x}{1+x^2}\,dx$ . Notice that the numerator $2x$ is exactly the derivative of $1+x^2$ . So:

\int \frac{2x}{1+x^2}\,dx = \log|1+x^2|

Since $1+x^2 > 0$ for all real $x$ , we drop the absolute value.

Hence:

\text{I.F.} = e^{\log(1+x^2)} = 1+x^2

Multiply the entire differential equation by the I.F.

(1+x^2)\frac{dy}{dx} + 2xy = \frac{1}{1+x^2}

The left-hand side is now exactly $\frac{d}{dx}\big[y(1+x^2)\big]$ . Why? Because:

\frac{d}{dx}\big[y(1+x^2)\big] = (1+x^2)\frac{dy}{dx} + y \cdot 2x

which matches perfectly.

Rewrite and integrate The equation becomes:

\frac{d}{dx}\big[y(1+x^2)\big] = \frac{1}{1+x^2}

Integrate both sides with respect to $x$ :

y(1+x^2) = \int \frac{1}{1+x^2}\,dx = \tan^{-1}x + c

where $c$ is the constant of integration. …

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Q.The solution of the differential equation $\frac{dy}{dx}+\frac{2xy}{1+x^2}=\frac{1}{(1+x^2)^2}$ is:
(A) $y(1+x^2)=c+\tan^{-1}x$
(B) $\frac{y}{1+x^2}=c+\tan^{-1}x$
(C) $y\log(1+x^2)=c+\tan^{-1}x$
(D) $y(1+x^2)=c+\sin^{-1}x$

Concept understanding — Integrating Factor Method

Integrating Factor Method

The Standard Form

The Integrating Factor

The Solution

A Quick Illustration

Method: Integrating factor of the form $e^{\log(\cdot)}$

Steps

Common Mistakes

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